Given the parallelogram $ABCD$ with the following properties:

• $AB = 3\sqrt{2}$
• $AD = 8$
• $\angle BAD = 45^\circ$
• Point $M$ divides $CD$ in the ratio $CM : MD = 1 : 2$
• Point $N$ divides $AD$ in the ratio $AN : ND = 3 : 1$

Let's find:

a) The Length of $AM$

To find $AM$, we first need to determine the coordinates of points $C$ and $D$. Since $ABCD$ is a parallelogram, we can set $A$ at $(0, 0)$, $B$ at $(3\sqrt{2}, 0)$, $D$ at $(0, 8)$, and $C$ at $(3\sqrt{2}, 8)$.

Coordinates of $M$:

Point $M$ divides $CD$ in the ratio $1:2$, so using the section formula, the coordinates of $M$ are:

$M = \left(\frac{2 \cdot 3\sqrt{2} + 1 \cdot 3\sqrt{2}}{1+2}, \frac{2 \cdot 8 + 1 \cdot 8}{1+2}\right) = \left(2\sqrt{2}, \frac{24}{3}\right) = (2\sqrt{2}, 8)$

Length of $AM$:

$AM = \sqrt{(2\sqrt{2} - 0)^2 + (8 - 0)^2} = \sqrt{8 + 64} = \sqrt{72} = 6\sqrt{2}$

b) The Length of $BN$

Coordinates of $N$:

Point $N$ divides $AD$ in the ratio $3:1$, so using the section formula, the coordinates of $N$ are:

$N = \left(\frac{3 \cdot 0 + 1 \cdot 0}{3+1}, \frac{3 \cdot 8 + 1 \cdot 0}{3+1}\right) = (0, 6)$

Length of $BN$:

$BN = \sqrt{(3\sqrt{2} - 0)^2 + (0 - 6)^2} = \sqrt{18 + 36} = \sqrt{54} = 3\sqrt{6}$

c) The Acute Angle Between $AM$ and $BN$

To find the acute angle between vectors $\overrightarrow{AM}$ and $\overrightarrow{BN}$, we use the dot product formula:

$\cos \theta = \frac{\overrightarrow{AM} \cdot \overrightarrow{BN}}{|\overrightarrow{AM}| \times |\overrightarrow{BN}|}$

Vectors:

$\overrightarrow{AM} = (2\sqrt{2}, 8) \quad \text{and} \quad \overrightarrow{BN} = (-3\sqrt{2}, -6)$

Dot Product:

$\overrightarrow{AM} \cdot \overrightarrow{BN} = (2\sqrt{2})(-3\sqrt{2}) + (8)(-6) = -12 - 48 = -60$

Magnitude of Vectors:

$|\overrightarrow{AM}| = 6\sqrt{2}, \quad |\overrightarrow{BN}| = 3\sqrt{6}$

Cosine of the Angle:

$\cos \theta = \frac{-60}{(6\sqrt{2})(3\sqrt{6})} = \frac{-60}{18\sqrt{12}} = \frac{-60}{36\sqrt{3}} = -\frac{5}{3\sqrt{3}} = -\frac{5\sqrt{3}}{9}$

Angle $\theta$:

$\theta = \cos^{-1}\left(-\frac{5\sqrt{3}}{9}\right)$

Since we need the acute angle, we consider the positive value:

$\theta = 180^\circ - \cos^{-1}\left(\frac{5\sqrt{3}}{9}\right)$

Would you like further details or calculations? Here are related questions you might find helpful:

1. How do you derive the section formula used here?
2. What is the geometric interpretation of the dot product of two vectors?
3. How would the problem change if $\angle BAD$ were different?
4. What is the significance of dividing a line segment in a given ratio?
5. How would you find coordinates in 3D using similar methods?

Tip: When working with ratios and coordinates, always use the section formula for accurate results!